# Finding the area between two functions.

• Jul 26th 2011, 01:18 AM
johnsy123
Finding the area between two functions.
Having problems with the following question.

For the equations y=sqrt{9-x} and y=x+3
a)Find the value of x when they intersect
b)Find the area between the curves from x=-1 to x=2.

-I found a to be x=0, but i am getting b wrong, weather its my arithmetic or my steps are wrong, i don't know.

-When inputting the two functions into an integral equation, should i take them away? multiply them? or......... i have been taking them away.

If anyone could help me with this, it would be greatly appreciated, thanks.
• Jul 26th 2011, 01:38 AM
FernandoRevilla
Re: Finding the area between two functions.
Quote:

Originally Posted by johnsy123
-When inputting the two functions into an integral equation, should i take them away? multiply them? or......... i have been taking them away.

$\displaystyle A=\int_{-1}^2|\sqrt{9-x}-(x+3)|\;dx=$

$\displaystyle \int_{-1}^0(\sqrt{9-x}-x-3)\;dx+\int_{0}^2(x+3-\sqrt{9-x})\;dx=\ldots$
• Jul 26th 2011, 05:11 AM
Prove It
Re: Finding the area between two functions.
http://i22.photobucket.com/albums/b3...n/integral.jpg

The area you are trying to evaluate is the combined pink and blue regions.

For the pink region, $\displaystyle \displaystyle y = \sqrt{9-x}$ is the greater function, so the area is $\displaystyle \displaystyle \int_{-1}^0{\sqrt{9-x}\,dx} - \int_{-1}^0{x + 3\,dx}$.

For the blue region, $\displaystyle \displaystyle y = x + 3$ is the greater function, so the area is $\displaystyle \displaystyle \int_0^2{x + 3\,dx} - \int_0^2{\sqrt{9-x}\,dx}$.